contestada

Gauss's law combines the electric field over a surface with the area of the surface. From Coulomb's law we know that the electric field falls off as 1/r2 of the distance r from the charge. How does the surface area change with r ?

Respuesta :

onyebuchinnaji

The change in surface area of Gaussian surface with radius (r) is 8πr.

Electric field from Coulomb's law

The electric field experienced by a charge is calculated as follows;

[tex]E = \frac{Q}{4\pi \varepsilon_o r^2}[/tex]

where;

  • E is the electric field
  • Q is the charge
  • r is the radius

The electric field reduces by a factor of [tex]\frac{1}{r^2}[/tex]

Surface area of a Gaussian surface;

The surface area of a sphere is given as;

[tex]A = 4\pi r^2[/tex]

Change in area with r

[tex]\frac{dA}{dr} = 8\pi r[/tex]

Thus, the change in surface area of Gaussian surface with radius (r) is 8πr.

Learn more about area of Gaussian surfaces here: https://brainly.com/question/17060446

Otras preguntas